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      <time title="创建时间：2023-01-23 05:01:31 / 修改时间：10:14:24" itemprop="dateCreated datePublished" datetime="2023-01-23T05:01:31+08:00">2023-01-23</time>
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        <h2 id="二叉树"><a href="#二叉树" class="headerlink" title="二叉树"></a>二叉树</h2><h3 id="二叉树的LCA（least-common-ancestor）"><a href="#二叉树的LCA（least-common-ancestor）" class="headerlink" title="二叉树的LCA（least-common-ancestor）"></a>二叉树的LCA（least-common-ancestor）</h3><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">import</span> java.util.*;</span><br><span class="line"></span><br><span class="line"><span class="keyword">public</span> <span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"></span><br><span class="line">    <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">lowestCommonAncestor</span> <span class="params">(TreeNode root, <span class="type">int</span> o1, <span class="type">int</span> o2)</span> &#123;</span><br><span class="line">        <span class="comment">// write code here</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">null</span>) <span class="keyword">return</span> -<span class="number">1</span>;</span><br><span class="line">        <span class="keyword">if</span>(root.val == o1) <span class="keyword">return</span> o1;</span><br><span class="line">        <span class="keyword">if</span>(root.val == o2) <span class="keyword">return</span> o2;</span><br><span class="line">        <span class="type">int</span> <span class="variable">l1</span> <span class="operator">=</span> lowestCommonAncestor(root.left, o1, o2);</span><br><span class="line">        <span class="type">int</span> <span class="variable">l2</span> <span class="operator">=</span> lowestCommonAncestor(root.right, o1, o2);</span><br><span class="line">        <span class="comment">// l1、l2 不可能同时为 -1</span></span><br><span class="line">        <span class="keyword">if</span>(l1 == -<span class="number">1</span>) <span class="keyword">return</span> l2;</span><br><span class="line">        <span class="keyword">if</span>(l2 == -<span class="number">1</span>) <span class="keyword">return</span> l1;</span><br><span class="line">        <span class="keyword">return</span> root.val;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>





<h3 id="二叉树的重建-层序遍历"><a href="#二叉树的重建-层序遍历" class="headerlink" title="二叉树的重建 + 层序遍历"></a>二叉树的重建 + 层序遍历</h3><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">import</span> java.util.*;</span><br><span class="line"></span><br><span class="line"><span class="keyword">public</span> <span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line">    <span class="comment">/**</span></span><br><span class="line"><span class="comment">     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可</span></span><br><span class="line"><span class="comment">     * 求二叉树的右视图</span></span><br><span class="line"><span class="comment">     * <span class="doctag">@param</span> xianxu int整型一维数组 先序遍历</span></span><br><span class="line"><span class="comment">     * <span class="doctag">@param</span> zhongxu int整型一维数组 中序遍历</span></span><br><span class="line"><span class="comment">     * <span class="doctag">@return</span> int整型一维数组</span></span><br><span class="line"><span class="comment">     */</span></span><br><span class="line">    <span class="keyword">static</span> <span class="keyword">class</span> <span class="title class_">Node</span> &#123;</span><br><span class="line">        <span class="type">int</span> val;</span><br><span class="line">        Node left, right;</span><br><span class="line">        Node(<span class="type">int</span> val) &#123;</span><br><span class="line">            <span class="built_in">this</span>.val = val;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    Node root;</span><br><span class="line">    <span class="type">int</span>[] pre, medi;</span><br><span class="line">    </span><br><span class="line">    Node <span class="title function_">build</span><span class="params">(<span class="type">int</span> l1, <span class="type">int</span> r1 ,<span class="type">int</span> l2, <span class="type">int</span> r2)</span> &#123;</span><br><span class="line">        <span class="keyword">if</span>(l1 &gt; r1) <span class="keyword">return</span> <span class="literal">null</span>;</span><br><span class="line">        <span class="type">int</span> <span class="variable">rootVal</span> <span class="operator">=</span> pre[l1];</span><br><span class="line">        <span class="type">Node</span> <span class="variable">node</span> <span class="operator">=</span> <span class="keyword">new</span> <span class="title class_">Node</span>(rootVal);</span><br><span class="line">        <span class="type">int</span> <span class="variable">idx</span> <span class="operator">=</span> -<span class="number">1</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> l2; i &lt;= r2; i++) &#123;</span><br><span class="line">            <span class="keyword">if</span>(medi[i] == rootVal) &#123;</span><br><span class="line">                idx = i;</span><br><span class="line">                <span class="keyword">break</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="type">int</span> <span class="variable">leftCnt</span> <span class="operator">=</span> idx - l2;</span><br><span class="line"></span><br><span class="line">        <span class="type">Node</span> <span class="variable">leftChild</span> <span class="operator">=</span> build(l1 + <span class="number">1</span>, l1 + leftCnt, l2, idx - <span class="number">1</span>);</span><br><span class="line">        <span class="type">Node</span> <span class="variable">rightChild</span> <span class="operator">=</span> build(l1 + leftCnt + <span class="number">1</span>, r1, idx + <span class="number">1</span>, r2);</span><br><span class="line">        node.left = leftChild;</span><br><span class="line">        node.right = rightChild;</span><br><span class="line">        <span class="keyword">return</span> node;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="keyword">public</span> <span class="type">int</span>[] solve (<span class="type">int</span>[] pre, <span class="type">int</span>[] medi) &#123;</span><br><span class="line">        <span class="comment">// write code here</span></span><br><span class="line">        <span class="built_in">this</span>.pre = pre;</span><br><span class="line">        <span class="built_in">this</span>.medi = medi;</span><br><span class="line">        <span class="type">int</span> <span class="variable">n</span> <span class="operator">=</span> pre.length;</span><br><span class="line">        </span><br><span class="line">        <span class="built_in">this</span>.root = build(<span class="number">0</span>, n - <span class="number">1</span>, <span class="number">0</span>, n - <span class="number">1</span>);</span><br><span class="line">        </span><br><span class="line">        List&lt;Integer&gt; res = <span class="keyword">new</span> <span class="title class_">ArrayList</span>&lt;&gt;();</span><br><span class="line">        Queue&lt;Node&gt; q = <span class="keyword">new</span> <span class="title class_">LinkedList</span>&lt;Node&gt;();</span><br><span class="line">        q.offer(root);</span><br><span class="line">        <span class="keyword">while</span>(q.size() &gt; <span class="number">0</span>) &#123;</span><br><span class="line">            <span class="type">int</span> <span class="variable">cnt</span> <span class="operator">=</span> q.size();</span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> <span class="number">0</span>; i &lt; cnt; i++) &#123;</span><br><span class="line">                <span class="type">Node</span> <span class="variable">node</span> <span class="operator">=</span> q.poll();</span><br><span class="line">                <span class="keyword">if</span>(i == <span class="number">0</span>) &#123;</span><br><span class="line">                    res.add(node.val);</span><br><span class="line">                &#125;</span><br><span class="line">                <span class="keyword">if</span>(node.right != <span class="literal">null</span>) q.offer(node.right);</span><br><span class="line">                <span class="keyword">if</span>(node.left != <span class="literal">null</span>) q.offer(node.left);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res.stream().mapToInt(Integer::valueOf).toArray();</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>



<h3 id="二叉树中的最大路径和"><a href="#二叉树中的最大路径和" class="headerlink" title="二叉树中的最大路径和"></a>二叉树中的最大路径和</h3><p>给定一颗二叉树，求二叉树的直径。<br>1.该题的直径定义为：树上任意两个节点路径长度的最大值；<br>2.该题路径长度定义为：不需要从根节点开始，也不需要在叶子节点结束，也不需要必须从父节点到子节点，一个节点到底另外一个节点走的边的数目；<br>3.这个路径可能穿过根节点，也可能不穿过；<br>4.树为空时，返回 0；</p>
<p>思路类似于<strong>树形DP</strong>求直径。</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">import</span> java.util.*;</span><br><span class="line"></span><br><span class="line"><span class="keyword">public</span> <span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line">    </span><br><span class="line">    <span class="type">int</span> <span class="variable">ans</span> <span class="operator">=</span> -(<span class="type">int</span>)<span class="number">1e8</span>;</span><br><span class="line">    <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">maxPathSum</span> <span class="params">(TreeNode root)</span> &#123;</span><br><span class="line">        <span class="comment">// write code here</span></span><br><span class="line">        dfs(root);</span><br><span class="line">        <span class="keyword">return</span> ans;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="type">int</span> <span class="title function_">dfs</span><span class="params">(TreeNode root)</span> &#123; <span class="comment">// return 从这个节点（此节点的值必选）出发的”链“的最大和</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">null</span>) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">        <span class="type">int</span> <span class="variable">lv</span> <span class="operator">=</span> dfs(root.left);</span><br><span class="line">        <span class="type">int</span> <span class="variable">rv</span> <span class="operator">=</span> dfs(root.right);</span><br><span class="line">        ans = Math.max(ans, (lv &gt; <span class="number">0</span> ? lv : <span class="number">0</span>) + (rv &gt; <span class="number">0</span> ? rv : <span class="number">0</span>) + root.val); <span class="comment">// 串上左右节点</span></span><br><span class="line">        <span class="type">int</span> <span class="variable">cs</span> <span class="operator">=</span> Math.max(lv, rv) &gt; <span class="number">0</span> ? Math.max(lv, rv) : <span class="number">0</span>; <span class="comment">// 向左出发、向右出发</span></span><br><span class="line">        <span class="keyword">return</span> cs + root.val; <span class="comment">// 本身的节点的值必选</span></span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
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